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Calculus Exam in 1 hour. New problem in OP

TL DRNot at all confident in his reflexive opinions of thingsRegistered Userregular

I do not know where the ln (0.994) bit comes from. Is it some special property of the derivative of a number ^10? Something to do with the chain rule? Thanks again. I'll be adding problems as they come along.

Second problem.

The average cost per item to produce q items is given by a(q) = q^2 - 90q +3500. Find the minimum value of the marginal cost. (Hint: First find the total cost C(q), and then MC(q). Then minimize MC(q).)

So to get C(q) I would just multiply the a(q) function by q, since (average cost per unit) times (number of units) = (total cost), right?
That gives me C(q) = q^3 - 90q + 3500q.
Marginal cost would be the derivative, correct? C'(q) = 3q^2 - 180q + 3500
I believe the next step is to set the function equal to zero to get the critical point(s).0 = 3q^2 - 180q + 35003500 = 3q^2 - 180q

No, it's not going to be ln(0.994), it's going to be ln(10), because d/dx of a^x is "(a^x)ln(x)", not "(a^x)ln(a)." I think. Honestly someone who has not gone years since taking calculus should be helping you, or at least someone who got an A when they took it years ago.

No, it's not going to be ln(0.994), it's going to be ln(10), because d/dx of a^x is "(a^x)ln(x)", not "(a^x)ln(a)." I think. Honestly someone who has not gone years since taking calculus should be helping you, or at least someone who got an A when they took it years ago.

Are you certain?

It seems to me that since a^x = e^[x ln(a)], it's far more likely that ln(a) will appear in the result than ln(x).

In fact, Wikipedia seems to think that the derivative is (a^x)ln(a), for a > 0.

P0 is the initial population at time zero (year 2000)
e is Euler's number (roughly 2.71, use e on your calculator for best results)
r is the rate (generally the unknown)
t is the time (10 years, referring to the period between 2000 and 2010)

Can you phrase the question this way? Otherwise I'm at a loss as to help you. I've only seen this type of question phrased like this.

I do not know where the ln (0.994) bit comes from. Is it some special property of the derivative of a number ^10? Something to do with the chain rule? Thanks again. I'll be adding problems as they come along.

Here's how I would do it (I know this is repeating and I think you know the right answer now, but I'll just write it out in hopes it solidifies it):

So the chain rule as I understand it is (derivative of outside function) (derivative of inside function) or f'(g(x)) times g'(x).

So I'd get something like
P(t) = 5.6(0.994)^t
P'(t) = 5.6 (0.994)^t ln(0.994)
Where ln(0.994) is the derivative of the ( )^10 function? Then substitute 10 and get
P'(10) = 5.6 (0.994)^10 ln(0.994)

This is correct.

You must have typed something into your calculator wrong to end up with 5.24. If you evaluate that correctly, you get -0.0317328642 (thanks google calculator). That's in millions of ppl/year so you need to multiply by a million to get it into people/year.

Marty81 on June 2009

0

TL DRNot at all confident in his reflexive opinions of thingsRegistered Userregular

The average cost per item to produce q items is given by a(q) = q^2 - 90q +3500. Find the minimum value of the marginal cost. (Hint: First find the total cost C(q), and then MC(q). Then minimize MC(q).)

So to get C(q) I would just multiply the a(q) function by q, since (average cost per unit) times (number of units) = (total cost), right?
That gives me C(q) = q^3 - 90q + 3500q.
Marginal cost would be the derivative, correct? C'(q) = 3q^2 - 180q + 3500
I believe the next step is to set the function equal to zero to get the critical point(s).0 = 3q^2 - 180q + 35003500 = 3q^2 - 180q

You need to differentiate again, then set C''(q) equal to 0. What you have is minimizing C, you need to minimize C'.

C''(q) = 6q - 180
q = 3
Then what, evaluate C'(3)?

Yes. Q=30 is the quantity, so to find the price plug back into your Marginal Cost function.

Hopefully you understand the general concept of minimizing functions (it seems like you mostly do but were a little confused here). If not, say so. But, to minimize C, you set C'=0. To minimize C', you set C''=0, etc.

EDIT: Q is actually 30. Think you just typed it wrong.

DJ-99 on June 2009

0

TL DRNot at all confident in his reflexive opinions of thingsRegistered Userregular

You need to differentiate again, then set C''(q) equal to 0. What you have is minimizing C, you need to minimize C'.

C''(q) = 6q - 180
q = 3
Then what, evaluate C'(3)?

Yes. Q=30 is the quantity, so to find the price plug back into your Marginal Cost function.

Hopefully you understand the general concept of minimizing functions (it seems like you mostly do but were a little confused here). If not, say so. But, to minimize C, you set C'=0. To minimize C', you set C''=0, etc.

EDIT: Q is actually 30. Think you just typed it wrong.

Yeah, twas just a typo. I got it. Thanks!

Next problem:
Minimize the cost of this fence. The area is 200 square feet, and the fencing along 3 sides costs $1/foot. The third side is $3/foot.

You need to differentiate again, then set C''(q) equal to 0. What you have is minimizing C, you need to minimize C'.

C''(q) = 6q - 180
q = 3
Then what, evaluate C'(3)?

Yes. Q=30 is the quantity, so to find the price plug back into your Marginal Cost function.

Hopefully you understand the general concept of minimizing functions (it seems like you mostly do but were a little confused here). If not, say so. But, to minimize C, you set C'=0. To minimize C', you set C''=0, etc.

EDIT: Q is actually 30. Think you just typed it wrong.

Yeah, twas just a typo. I got it. Thanks!

Next problem:
Minimize the cost of this fence. The area is 200 square feet, and the fencing along 3 sides costs $1/foot. The third side is $3/foot.

Good luck. If you have any spare time at the end just go back and double-check all your arithmetic. But, it is calculus, so understanding the concepts should at least gain you a bunch of points, even if you make a multiplication error.

DJ-99 on June 2009

0

TL DRNot at all confident in his reflexive opinions of thingsRegistered Userregular